3.7 KiB
a) Choose a number between 51 and 70. This will be number A. Choose another number between 80 and 120. This will be number B. Using 2’s complement 8-bit binary arithmetic, calculate number A – number B giving your answer in binary and decimal. (5 marks)
Number A - 64 - 0100 0000 Number B - 96 - 0110 0000
| Binary Subtraction | |
|---|---|
| Num A | 0100 0000 |
| Num B | 0110 0000 |
| Flip B | 1001 1111 |
| Add 1 | 1010 0000 |
| Num A | 0100 0000 |
| Num B | 1010 0000 |
| Output | 1110 0000 |
| 1s Comp | 0001 1111 |
| 2s Comp | -0010 0000 |
| Decimal | -32 |
b) Choose one number from the set { 30, 31, 33, 34, 35, 35, 37, 38, 39 } this is numberC. Choose a number from { 3, 5, 6 } this is numberD. Using 8-bit binary arithmetic, calculate numberC × numberD giving your answer in binary and decimal.
Number C - 34 - 0010 0010 Number D - 6 - 0000 0110 - 2^1, 2^2
| Binary Multiplication | |
|---|---|
| Num C | 0010 0010 |
| Mult 2^1 | 0100 0100 |
| Num C | 0010 0010 |
| Mult 2^2 | 1000 1000 |
| Add | 1100 1100 |
| Decimal | 204 |
Here, we split our exponent into 2, giving a value of 2 and 4 (2^2), so we can multiply the mantissa by both, and add them together.
c) Using numberC and numberD from (b), use binary arithmetic to calculate: numberC.125 + numberD.375 giving your answer in binary and decimal. Convert numberC.125 into IEEE-754 format.
| numberC | 0010 0010 |
|---|---|
| Floating Point | 0010 0010 . 001 |
| C.125 | 100010.001 |
| Decimal | 34.125 |
| numberD | 0000 0010 |
|---|---|
| Floating Point | 0000 0110 . 011 |
| D.375 | 110.011 |
| Decimal | 6.375 |
| Binary Addition | |
|---|---|
| Num C.125 | 0010 0010.001 |
| Num D.375 | 0000 0110.011 |
| Add | 0010 1000.100 |
| Decimal | 40.5 |
Here, we add the floating points to the two numbers, and convert both to decimal to check our answer. Adding them together in the 3rd table to get an answer in both decimal and binary.
| Convert -> Floating Point | Binary | Decimal |
|---|---|---|
| Num C.125 | 100010.001 | 34.125 |
| Normalised | 1.00010001 x 2^5 | 34.125 |
| Remove leading 1 | 0.00010001 | 0.06640625 |
| Sign | 0 (Positive) | 0 |
| Exponent | 10000100 | 132 (2^5) |
| Mantissa | 00010001000000000000000 | 1.06640625 |
| Total | 01000010000010001000000000000000 | 2^5 x 1.06640625 |
Here, since the sign is positive, it takes the value of 0. The exponent is 2^5, but in IEEE-754, you would take the exponent from the normalised value and add to the bias (127 in 32bit). I then got the mantissa by removing the leading 1, since it is assumed, from the normalised value, and adding the trailing 0's to get 23 bits. Concatenating these values I then get the final answer in decimal and binary.
d) Choose a number between 140 and 160 - this is numberE. Convert numberE to hexadecimal. Choose a number between 170 and 255 - this is numberF. Convert numberF into octal. Using 8-bit binary, calculate numberE OR numberF. Give your answer in binary, octal, decimal and hexadecimal.
NumberE - 160 NumberF - 223
| NumberE | 160 |
|---|---|
| Binary | 0b10100000 |
| Hexadecimal | 0xA0 |
| NumberF | 223 |
|---|---|
| /8 | 27r7 |
| /8 | 3r3 |
| /8 | 0r3 |
| Octal | 0337 |
| Binary | 1101 1111 |
| NumE OR NumF | |
|---|---|
| NumE | 0b1010 0000 |
| NumF | 0b1101 1111 |
| OR (Binary) | 0b11111111 |
| OR (Decimal) | 255 |
| OR (Hexadecimal) | 0xFF |
| OR (Octal) | 0377 |
Here, I have picked two numbers, the first of which is converted to hexadecimal, and the second converted to octal using a division conversion method on the decimal value. Then I have done a bitwise operation for OR on the two binary values to get a binary value, and then converted to decimal, hexadecimal and octal - in this case it is the maximum value for each using 8 bit binary.